A count is correct only if every valid outcome is counted exactly once. Too many routes to the same outcome gives over-counting, and a missing case gives under-counting.
This lesson finishes the permutations and combinations chapter. It builds on the shortcut warned about in counting selections with required members.
How do I test for over-counting?
Choose one outcome and count how many times your method produces it. If it is more than once, your method is wrong. The fix is usually to split into separate cases, or to divide by the number of repeats.
Worked example: why 210 is wrong
A team of 5 is chosen from 5 boys and 4 girls with at least 2 girls. The shortcut is 4C2 × 7C3 = 6 × 35 = 210. The correct answer, found by cases earlier, is 105.
Test a team with exactly 3 girls, say girls G1, G2 and G3 and boys B1 and B2. The method “pick 2 girls first” can start with G1 and G2, G1 and G3, or G2 and G3. That is 3 routes to one team.
Now count the multiplicity for every type of team:
| Girls in the team | Teams | Times counted | Contribution |
|---|---|---|---|
| 2 | 60 | 1 | 60 |
| 3 | 40 | 3 | 120 |
| 4 | 5 | 6 | 30 |
The sum is 60 + 120 + 30 = 210, which exactly explains the wrong answer. The real total is 60 + 40 + 5 = 105.
Worked example: repeated letters
How many arrangements are there of the letters of BANANA?
There are 6 letters, but A appears 3 times and N appears 2 times. The count 6! = 720 treats every A as different, and every N as different.
Each real arrangement is counted 3! × 2! = 6 × 2 = 12 times. So the answer is 720 ÷ 12 = 60.
How do I test for under-counting?
List the valid cases and tick each one off. In “at least 2 girls” out of 4 girls, the cases are 2, 3 and 4 girls. If your working has only 2 and 3, you are missing a case.
A second test is the complement. If cases and complement agree, as they did at 105, both are very likely right.
The mistake that costs marks
It is easy to stop once a method produces a clean number. The 210 above looks tidy, and nothing in the working flags it.
Build the habit of one check per question: test a tiny version, use a second method, or ask how many ways each outcome is produced.
Check yourself
How many different arrangements are there of the letters of SEVEN?
Answer
There are 5 letters, with E appearing twice.
5! = 120 counts each arrangement twice, once for each order of the two E’s.
The answer is 120 ÷ 2! = 60.
Check with a smaller case: the word EE has 1 arrangement, and 2! ÷ 2! = 1.
What to study next
Test all four skills on the permutations and combinations practice set. Record your slips in the mistake log.
If you want a teacher to question your method on real problems, see online one-to-one Additional Mathematics tuition.